Question: Rewrite the function by completing the square. $h(x)= 4 x^{2} -36 x +81$ $h(x)=$
Explanation: $\begin{aligned} h(x)&=4 x^2 -36 x +81 \\\\ &=4 \left(x^2 -9 x\right) +81 \end{aligned}$ Now we want to complete $x^2 -9 x$ into a perfect square. To do that, we should add $\left(\dfrac{{-9}}{2}\right)^2={\dfrac{81}{4}}$ to it: $x^2{-9}x+{\dfrac{81}{4}}=\left(x -\dfrac{9}{2}\right)^2$ We add ${\dfrac{81}{4}}$ inside the parentheses, and subtract ${4}\cdot{\dfrac{81}{4}}$ outside them, to keep the expression equivalent. $\begin{aligned} &\phantom{=}{4} \left(x^2 -9 x\right) +81 \\\\ &={4}\left(x^2 -9 x+{\dfrac{81}{4}}\right) +81 -{4}\cdot{\dfrac{81}{4}} \\\\ &=4 \left(x -\dfrac{9}{2}\right)^2 +81 -81 \\\\ &=4 \left(x -\dfrac{9}{2}\right)^2 +0 \end{aligned}$ In conclusion, the function after completing the square is written as: $h(x)=4 \left(x -\dfrac{9}{2}\right)^2 +0$